Cho \(a,b,c\) thỏa mãn \(\dfrac{a^3-1}{a}=\dfrac{b^3-1}{b}=\dfrac{c^3-1}{c}\)
CMR \(abc+1=0\)
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DH
1
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2021
\(\dfrac{a^3}{1+b}+\dfrac{1+b}{4}+\dfrac{1}{2}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)}{8\left(a+b\right)}}=\dfrac{3a}{2}\)
\(\dfrac{b^3}{1+c}+\dfrac{1+c}{4}+\dfrac{1}{2}\ge\dfrac{3b}{2}\) ; \(\dfrac{c^3}{1+a}+\dfrac{1+a}{4}+\dfrac{1}{2}\ge\dfrac{3c}{2}\)
\(\Rightarrow VT+\dfrac{a+b+c}{4}+\dfrac{9}{4}\ge\dfrac{3}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{5}{4}\left(a+b+c\right)-\dfrac{9}{4}\ge\dfrac{5}{4}.3\sqrt[3]{abc}-\dfrac{9}{4}=\dfrac{3}{2}\)
HT
1
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 8 2021
Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)
\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)
3 tháng 4 2022
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
24 tháng 6 2021
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
AT
24 tháng 6 2021
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 6 2021
Bài này đã có ở đây:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24
Cho $a=b=c=1$ thì thỏa mãn đẳng thức nhưng $abc+1=2\neq 0$
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